use strict;
use warnings;

my $find    = "20150518";
my $replace = "18 May 2015";

while (my $line = <DATA>) {
    $line =~ s/$find/$replace/g;
    print $line;
}

__DATA__
20150518
[download]

Prints:

18 May 2015
[download]

as expected which leads me to think that the code you show is not the code you executed to get the error. Maybe you can modify the test code I've given to reproduce your error then show us that version?

Ya exactly its printing on the terminal correctly but the actual data not changing. But i want to change in the file also.

local $^I = '.old';
local @ARGV = ( 'your_file_name' );
my $find = "20150518";
my $replace = "18 May 2015";
while ( <>){
    s/$find/$replace/g;
        print;
}
[download]

Is it possible to change the data without changing extension, name & path of the file????

Nope, you essentially can't change a free-format sequential file like that. That's why, behind the scene, you have to rename your original file (say into file.old), and then write your changed content to a new file having the name of your original file.

The alternative is to read the whole file into memory, make the changes in memory and then clobber the content of the original file and write the new content to it.

The modified file will have same extension and name as original file
Original/unmodified file will get new extension *.old.

You can open the file for both read and write, but that can be tricky. Better to create a new file and rename to old file name.

use strict;
use warnings;
use File::Copy;

my $infilename = 'Pull-2015-08-15-23.txt';
my $outfilename = 'new.file';
my $find = '20150518';
my $replace = '18 May 2015';

open $infile,'<',$infilename;
open $outfile,'>',$outfilename;
while (my $line = <$infile>) {
  $line =~ s/$find/$replace/;
  print $outfile $line;
}
close $infile;
close $outfile;
move($outfilename,$infilename);
[download]

You may find modules like Date::Parse and Date::Format useful to deal with dates in general, and not just this specific example.

Update:

use Date::Parse;
use Date::Format;

@time = strptime("20150518");
print strftime("%e %b %Y",@time);
[download]

# #1137654

my $find = "20150518<";          # assumes that a date-like value not 
+followed by a '<' is not what you seek
my $replace = "18 May 2015";
while(my $line = <DATA>)
{
      if($line =~ $find) {    # select lines to print, unless you want
+ all lines * Note1
        s/$find/$replace/g;  # UPDATE: but the slash-g is utterly unne
+cessary 
            print $line;         # as it would act only on single line
+s in this case
    }
}

# * Note1: acknowledged: you want all lines.  (another update)


=head output

20150518<114319747011000001>,
20150518<bulkbull3>,


=cut


__DATA__
20150518<114319747011000001>,
20150518230002,< >,
917696706652,01,
,
52727,
rch,,
admin,,
01,21314,,
bulkpull2,,
20150518<bulkbull3>,
[download]

You can't open DATA directly like that, which is what the error message is telling you. Either change DATA to something else, or (preferably) switch to the three form method of opening files: open my $data, '<', 'Pull-2015-5-18-23.txt' or die 'failed to open file';

Did you run the OP's code? Did you test your assumption? Obviously not because:

use strict;
use warnings;

my $filename = 'data.txt';

open DATA, '>', $filename or die "Can't create $filename: $!\n";

print DATA <<STUFF;
Stuff written to disk using file handle DATA
STUFF

close DATA;

open my $inFile, '<', 'data.txt' or die "Can't open $filename: $!\n";
print <$inFile>;
close $inFile;
[download]

Prints:

Stuff written to disk using file handle DATA
[download]

which would fail with the OP's error message if you were right. In fact I can't see how you get your implied error of "Can't use DATA in a file open statement" from 'Can't modify constant item in scalar assignment at date.pl line 1, near "10,"'.

Premature optimization is the root of all job security